3.2.68 \(\int (c+a^2 c x^2)^3 \text {ArcTan}(a x) \, dx\) [168]

Optimal. Leaf size=161 \[ -\frac {4 c^3 \left (1+a^2 x^2\right )}{35 a}-\frac {3 c^3 \left (1+a^2 x^2\right )^2}{70 a}-\frac {c^3 \left (1+a^2 x^2\right )^3}{42 a}+\frac {16}{35} c^3 x \text {ArcTan}(a x)+\frac {8}{35} c^3 x \left (1+a^2 x^2\right ) \text {ArcTan}(a x)+\frac {6}{35} c^3 x \left (1+a^2 x^2\right )^2 \text {ArcTan}(a x)+\frac {1}{7} c^3 x \left (1+a^2 x^2\right )^3 \text {ArcTan}(a x)-\frac {8 c^3 \log \left (1+a^2 x^2\right )}{35 a} \]

[Out]

-4/35*c^3*(a^2*x^2+1)/a-3/70*c^3*(a^2*x^2+1)^2/a-1/42*c^3*(a^2*x^2+1)^3/a+16/35*c^3*x*arctan(a*x)+8/35*c^3*x*(
a^2*x^2+1)*arctan(a*x)+6/35*c^3*x*(a^2*x^2+1)^2*arctan(a*x)+1/7*c^3*x*(a^2*x^2+1)^3*arctan(a*x)-8/35*c^3*ln(a^
2*x^2+1)/a

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Rubi [A]
time = 0.05, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4998, 4930, 266} \begin {gather*} \frac {1}{7} c^3 x \left (a^2 x^2+1\right )^3 \text {ArcTan}(a x)+\frac {6}{35} c^3 x \left (a^2 x^2+1\right )^2 \text {ArcTan}(a x)+\frac {8}{35} c^3 x \left (a^2 x^2+1\right ) \text {ArcTan}(a x)-\frac {c^3 \left (a^2 x^2+1\right )^3}{42 a}-\frac {3 c^3 \left (a^2 x^2+1\right )^2}{70 a}-\frac {4 c^3 \left (a^2 x^2+1\right )}{35 a}-\frac {8 c^3 \log \left (a^2 x^2+1\right )}{35 a}+\frac {16}{35} c^3 x \text {ArcTan}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)^3*ArcTan[a*x],x]

[Out]

(-4*c^3*(1 + a^2*x^2))/(35*a) - (3*c^3*(1 + a^2*x^2)^2)/(70*a) - (c^3*(1 + a^2*x^2)^3)/(42*a) + (16*c^3*x*ArcT
an[a*x])/35 + (8*c^3*x*(1 + a^2*x^2)*ArcTan[a*x])/35 + (6*c^3*x*(1 + a^2*x^2)^2*ArcTan[a*x])/35 + (c^3*x*(1 +
a^2*x^2)^3*ArcTan[a*x])/7 - (8*c^3*Log[1 + a^2*x^2])/(35*a)

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 4998

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(-b)*((d + e*x^2)^q/(2*c
*q*(2*q + 1))), x] + (Dist[2*d*(q/(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x]), x], x] + Simp[x*(d
+ e*x^2)^q*((a + b*ArcTan[c*x])/(2*q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[q, 0]

Rubi steps

\begin {align*} \int \left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x) \, dx &=-\frac {c^3 \left (1+a^2 x^2\right )^3}{42 a}+\frac {1}{7} c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)+\frac {1}{7} (6 c) \int \left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x) \, dx\\ &=-\frac {3 c^3 \left (1+a^2 x^2\right )^2}{70 a}-\frac {c^3 \left (1+a^2 x^2\right )^3}{42 a}+\frac {6}{35} c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac {1}{7} c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)+\frac {1}{35} \left (24 c^2\right ) \int \left (c+a^2 c x^2\right ) \tan ^{-1}(a x) \, dx\\ &=-\frac {4 c^3 \left (1+a^2 x^2\right )}{35 a}-\frac {3 c^3 \left (1+a^2 x^2\right )^2}{70 a}-\frac {c^3 \left (1+a^2 x^2\right )^3}{42 a}+\frac {8}{35} c^3 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac {6}{35} c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac {1}{7} c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)+\frac {1}{35} \left (16 c^3\right ) \int \tan ^{-1}(a x) \, dx\\ &=-\frac {4 c^3 \left (1+a^2 x^2\right )}{35 a}-\frac {3 c^3 \left (1+a^2 x^2\right )^2}{70 a}-\frac {c^3 \left (1+a^2 x^2\right )^3}{42 a}+\frac {16}{35} c^3 x \tan ^{-1}(a x)+\frac {8}{35} c^3 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac {6}{35} c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac {1}{7} c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)-\frac {1}{35} \left (16 a c^3\right ) \int \frac {x}{1+a^2 x^2} \, dx\\ &=-\frac {4 c^3 \left (1+a^2 x^2\right )}{35 a}-\frac {3 c^3 \left (1+a^2 x^2\right )^2}{70 a}-\frac {c^3 \left (1+a^2 x^2\right )^3}{42 a}+\frac {16}{35} c^3 x \tan ^{-1}(a x)+\frac {8}{35} c^3 x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)+\frac {6}{35} c^3 x \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)+\frac {1}{7} c^3 x \left (1+a^2 x^2\right )^3 \tan ^{-1}(a x)-\frac {8 c^3 \log \left (1+a^2 x^2\right )}{35 a}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 83, normalized size = 0.52 \begin {gather*} \frac {c^3 \left (-a^2 x^2 \left (57+24 a^2 x^2+5 a^4 x^4\right )+6 a x \left (35+35 a^2 x^2+21 a^4 x^4+5 a^6 x^6\right ) \text {ArcTan}(a x)-48 \log \left (1+a^2 x^2\right )\right )}{210 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + a^2*c*x^2)^3*ArcTan[a*x],x]

[Out]

(c^3*(-(a^2*x^2*(57 + 24*a^2*x^2 + 5*a^4*x^4)) + 6*a*x*(35 + 35*a^2*x^2 + 21*a^4*x^4 + 5*a^6*x^6)*ArcTan[a*x]
- 48*Log[1 + a^2*x^2]))/(210*a)

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Maple [A]
time = 0.09, size = 102, normalized size = 0.63

method	result	size
derivativedivides	\(\frac {\frac {c^{3} \arctan \left (a x \right ) a^{7} x^{7}}{7}+\frac {3 a^{5} c^{3} x^{5} \arctan \left (a x \right )}{5}+a^{3} c^{3} x^{3} \arctan \left (a x \right )+a \,c^{3} x \arctan \left (a x \right )-\frac {c^{3} \left (\frac {5 a^{6} x^{6}}{6}+4 a^{4} x^{4}+\frac {19 a^{2} x^{2}}{2}+8 \ln \left (a^{2} x^{2}+1\right )\right )}{35}}{a}\)	\(102\)
default	\(\frac {\frac {c^{3} \arctan \left (a x \right ) a^{7} x^{7}}{7}+\frac {3 a^{5} c^{3} x^{5} \arctan \left (a x \right )}{5}+a^{3} c^{3} x^{3} \arctan \left (a x \right )+a \,c^{3} x \arctan \left (a x \right )-\frac {c^{3} \left (\frac {5 a^{6} x^{6}}{6}+4 a^{4} x^{4}+\frac {19 a^{2} x^{2}}{2}+8 \ln \left (a^{2} x^{2}+1\right )\right )}{35}}{a}\)	\(102\)
risch	\(-\frac {i c^{3} x \left (5 a^{6} x^{6}+21 a^{4} x^{4}+35 a^{2} x^{2}+35\right ) \ln \left (i a x +1\right )}{70}+\frac {i c^{3} a^{6} x^{7} \ln \left (-i a x +1\right )}{14}-\frac {a^{5} c^{3} x^{6}}{42}+\frac {3 i c^{3} a^{4} x^{5} \ln \left (-i a x +1\right )}{10}-\frac {4 a^{3} c^{3} x^{4}}{35}+\frac {i c^{3} a^{2} x^{3} \ln \left (-i a x +1\right )}{2}-\frac {19 a \,c^{3} x^{2}}{70}+\frac {i c^{3} x \ln \left (-i a x +1\right )}{2}-\frac {8 c^{3} \ln \left (-a^{2} x^{2}-1\right )}{35 a}\)	\(168\)
meijerg	\(\frac {c^{3} \left (-\frac {a^{2} x^{2} \left (4 a^{4} x^{4}-6 a^{2} x^{2}+12\right )}{42}+\frac {4 a^{8} x^{8} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{7 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{7}\right )}{4 a}+\frac {3 c^{3} \left (\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+6\right )}{15}+\frac {4 a^{6} x^{6} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (a^{2} x^{2}+1\right )}{5}\right )}{4 a}+\frac {3 c^{3} \left (-\frac {2 a^{2} x^{2}}{3}+\frac {4 a^{4} x^{4} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}\right )}{4 a}+\frac {c^{3} \left (\frac {4 a^{2} x^{2} \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{4 a}\)	\(246\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^3*arctan(a*x),x,method=_RETURNVERBOSE)

[Out]

1/a*(1/7*c^3*arctan(a*x)*a^7*x^7+3/5*a^5*c^3*x^5*arctan(a*x)+a^3*c^3*x^3*arctan(a*x)+a*c^3*x*arctan(a*x)-1/35*
c^3*(5/6*a^6*x^6+4*a^4*x^4+19/2*a^2*x^2+8*ln(a^2*x^2+1)))

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Maxima [A]
time = 0.25, size = 99, normalized size = 0.61 \begin {gather*} -\frac {1}{210} \, {\left (5 \, a^{4} c^{3} x^{6} + 24 \, a^{2} c^{3} x^{4} + 57 \, c^{3} x^{2} + \frac {48 \, c^{3} \log \left (a^{2} x^{2} + 1\right )}{a^{2}}\right )} a + \frac {1}{35} \, {\left (5 \, a^{6} c^{3} x^{7} + 21 \, a^{4} c^{3} x^{5} + 35 \, a^{2} c^{3} x^{3} + 35 \, c^{3} x\right )} \arctan \left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x),x, algorithm="maxima")

[Out]

-1/210*(5*a^4*c^3*x^6 + 24*a^2*c^3*x^4 + 57*c^3*x^2 + 48*c^3*log(a^2*x^2 + 1)/a^2)*a + 1/35*(5*a^6*c^3*x^7 + 2
1*a^4*c^3*x^5 + 35*a^2*c^3*x^3 + 35*c^3*x)*arctan(a*x)

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Fricas [A]
time = 3.57, size = 101, normalized size = 0.63 \begin {gather*} -\frac {5 \, a^{6} c^{3} x^{6} + 24 \, a^{4} c^{3} x^{4} + 57 \, a^{2} c^{3} x^{2} + 48 \, c^{3} \log \left (a^{2} x^{2} + 1\right ) - 6 \, {\left (5 \, a^{7} c^{3} x^{7} + 21 \, a^{5} c^{3} x^{5} + 35 \, a^{3} c^{3} x^{3} + 35 \, a c^{3} x\right )} \arctan \left (a x\right )}{210 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x),x, algorithm="fricas")

[Out]

-1/210*(5*a^6*c^3*x^6 + 24*a^4*c^3*x^4 + 57*a^2*c^3*x^2 + 48*c^3*log(a^2*x^2 + 1) - 6*(5*a^7*c^3*x^7 + 21*a^5*
c^3*x^5 + 35*a^3*c^3*x^3 + 35*a*c^3*x)*arctan(a*x))/a

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Sympy [A]
time = 0.41, size = 117, normalized size = 0.73 \begin {gather*} \begin {cases} \frac {a^{6} c^{3} x^{7} \operatorname {atan}{\left (a x \right )}}{7} - \frac {a^{5} c^{3} x^{6}}{42} + \frac {3 a^{4} c^{3} x^{5} \operatorname {atan}{\left (a x \right )}}{5} - \frac {4 a^{3} c^{3} x^{4}}{35} + a^{2} c^{3} x^{3} \operatorname {atan}{\left (a x \right )} - \frac {19 a c^{3} x^{2}}{70} + c^{3} x \operatorname {atan}{\left (a x \right )} - \frac {8 c^{3} \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{35 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**3*atan(a*x),x)

[Out]

Piecewise((a**6*c**3*x**7*atan(a*x)/7 - a**5*c**3*x**6/42 + 3*a**4*c**3*x**5*atan(a*x)/5 - 4*a**3*c**3*x**4/35
+ a**2*c**3*x**3*atan(a*x) - 19*a*c**3*x**2/70 + c**3*x*atan(a*x) - 8*c**3*log(x**2 + a**(-2))/(35*a), Ne(a,
0)), (0, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^3*arctan(a*x),x, algorithm="giac")

[Out]

sage0*x

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Mupad [B]
time = 0.25, size = 89, normalized size = 0.55 \begin {gather*} -\frac {c^3\,\left (48\,\ln \left (a^2\,x^2+1\right )+57\,a^2\,x^2+24\,a^4\,x^4+5\,a^6\,x^6-210\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )-126\,a^5\,x^5\,\mathrm {atan}\left (a\,x\right )-30\,a^7\,x^7\,\mathrm {atan}\left (a\,x\right )-210\,a\,x\,\mathrm {atan}\left (a\,x\right )\right )}{210\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)*(c + a^2*c*x^2)^3,x)

[Out]

-(c^3*(48*log(a^2*x^2 + 1) + 57*a^2*x^2 + 24*a^4*x^4 + 5*a^6*x^6 - 210*a^3*x^3*atan(a*x) - 126*a^5*x^5*atan(a*
x) - 30*a^7*x^7*atan(a*x) - 210*a*x*atan(a*x)))/(210*a)

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